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I am currently trying to understand the proof of Theorem 4 in John G. Heywood's paper On Uniqueness Questions in the Theory of Viscous Flow. Near the end of the proof, the inequality $ (\nabla u)^2 \geq \left( \frac{1}{r} \partial_\theta u_r \right)^2 $ is stated, and I am currently trying to understand why this inequality holds. After expanding the expression for $ |\nabla u|^2 $ to get $$ |\nabla u|^2 = (\partial_r u_r)^2 + (\partial_r u_\theta)^2 + \frac{1}{r^2} (\partial_\theta u_r - u_\theta)^2 + \frac{1}{r^2} (\partial_\theta u_\theta + u_r)^2, $$ the inequality seems less clear to me. Since $ \frac{1}{r^2} (\partial_\theta u_r - u_\theta)^2 = \left( \frac{1}{r} \partial_\theta u_r \right)^2 - \frac{2}{r^2} u_\theta \partial_\theta u_r + \frac{1}{r^2} u_\theta^2 $, the desired inequality $ |\nabla u|^2 - \left( \frac{1}{r} \partial_\theta u_r \right) \geq 0 $ is equivalent to $$ \left( (\partial_r u_r)^2 + (\partial_r u_\theta)^2 + \frac{1}{r^2} (\partial_\theta u_\theta + u_r)^2 + \frac{1}{r^2} u_\theta^2 \right) - \frac{2}{r^2} u_\theta \partial_\theta u_r \geq 0. $$ This inequality doesn't seem to hold for arbitrary functions $ u(r, \theta) $. For instance, if we take $ u(r, \theta) = (u_r, u_\theta) = (\theta, 1) $, then the above inequality reduces to $ \theta^2 \geq 1 $, and this doesn't hold for $ \theta \in (1, 2 \pi) $. This doesn't take into account some potentially relevant information about $ u(r,\theta) $ in the context of Heywood's proof, such as:

  1. The divergence-free condition $ \nabla \cdot u = 0 $, which in polar coordinates is $$ \partial_r u_r + \frac{1}{r} \partial_\theta u_\theta + \frac{1}{r} u_r = 0. $$

  2. The no-slip boundary condition and the decay-at-infinity condition, which state $$ u(r, \theta) \Large\mid_{\normalsize{r=1}} \normalsize = 0, \qquad \lim_{r \to \infty} u(r, \theta) = 0 \qquad \forall \theta \in [0,2\pi). $$

  3. The Fourier expansion for the radial function $ u_r (r, \theta) $ obtained earlier in the proof, $$ u_r (r, \theta) = a_0 (r) + \sum_{n=1}^\infty (a_n (r) \cos{(n \theta)} + b_n(r) \sin{(n \theta)}). $$

  4. The finite-energy condition $ \int_\Omega |\nabla u|^2 < \infty $, which in our case takes the form $$ \int_0^{2 \pi} \int_1^\infty \left( (\partial_r u_r )^2 + (\partial_r u_\theta)^2 + \frac{1}{r^2} (\partial_\theta u_r - u_\theta)^2 + \frac{1}{r^2} (\partial_\theta u_\theta + u_r)^2 \right) \hspace{0.5 mm} r \hspace{0.5 mm} dr \hspace{0.5 mm} d\theta < \infty. $$

First of all, is Heywood's inequality true? Does it follow from any of the above conditions on $ u(r, \theta) $ listed above? I tried using the substitution $ u_r = - \partial_\theta u_\theta - r \partial_r u_r $ from the divergence-free condition, which simplifies the inequality to $$ \left( 2 (\partial_r u_r)^2 + ( \partial_r u_\theta )^2 + \frac{1}{r^2} u_\theta^2 \right) - \frac{2}{r^2} u_\theta \partial_\theta u_r \geq 0. $$ However, I am unsure how to proceed from here. I have been testing divergence-free functions of the form $ u(r, \theta) = (u_r, u_\theta) = \left( \frac{1}{r} \partial_\theta \psi, - \partial_r \psi \right) $ to try and find a counterexample because I'm not sure if the inequality follows from the divergence-free condition or not. Any help would be appreciated.

Edit: Here are some of the details for how I derived $ |\nabla u|^2 $ in the post. First, let $$ g_{ij} = \begin{bmatrix} 1 & 0 \\ 0 & r^2 \end{bmatrix} $$ be the metric tensor in Euclidean polar coordinates. Let $ v = v^r \partial_r + v^\theta \partial_\theta $, then $$ \begin{split} |\nabla v|_g^2 & = g(\nabla v, \nabla v) = g^{ik} g^{j \ell} (\nabla_i v_j) (\nabla_k v_\ell) \\ & = (\nabla_r v_r)^2 + \frac{1}{r^2} \left( ( \nabla_r v_\theta )^2 + (\nabla_\theta v_r)^2 \right) + \frac{1}{r^4} (\nabla_\theta v_\theta)^2. \end{split} $$ Using the formula for the covariant derivative $ \nabla_i v_j = \partial_i v_j - \Gamma_{ij}^k v_k $ (where Christoffel symbols are all zero except for $ \Gamma_{\theta \theta}^r = -r $ and $ \Gamma_{r \theta}^\theta = \Gamma_{\theta r}^\theta = 1/r $) we get $$ \begin{split} (\nabla_r v_r)^2 & = \left( \partial_r v_r \right)^2, \\ (\nabla_r v_\theta)^2 & = \left( \partial_r v_\theta - \Gamma_{r \theta}^\theta v_\theta \right)^2 = \left( \partial_r v_\theta - \frac{1}{r} v_\theta \right)^2, \\ (\nabla_\theta v_r)^2 & = \left( \partial_\theta v_r - \Gamma_{\theta r}^\theta v_\theta \right)^2 = \left( \partial_\theta v_r - \frac{1}{r} v_\theta \right)^2, \\ (\nabla_\theta v_\theta)^2 & = \left( \partial_\theta v_\theta - \Gamma_{\theta \theta}^r v_r \right)^2 = \left( \partial_\theta v_\theta + r \right)^2. \end{split} $$ We now convert to the frame $ u = u_r E_r + u_\theta E_\theta $ using the transformation $ (v_r, v_\theta) = (u_r, r u_\theta) $. This follows from $ u_r = v_r $ and $$ u_\theta = r u^\theta = r g^{\theta \theta} v_\theta = r \left( \frac{1}{r^2} \right) v_\theta = \frac{1}{r} v_\theta. $$ Recomputing each $ (\nabla_i v_j)^2 $ with this transformation gives us $$ \begin{split} (\nabla_r v_r)^2 & = (\partial_r u_r)^2, \\ (\nabla_r v_\theta)^2 & = \left( \partial_r (r u_\theta) - \frac{1}{r} (r u_\theta) \right)^2 = \left( r \partial_r u_\theta \right)^2, \\ (\nabla_\theta v_r)^2 & = \left( \partial_\theta u_r - \frac{1}{r} (r u_\theta) \right)^2 = \left( \partial_\theta u_r - u_\theta \right)^2, \\ (\nabla_\theta v_\theta)^2 & = \left( \partial_\theta (r u_\theta) + r u_r \right)^2 = \left( r \partial_\theta u_\theta + ru_r \right)^2. \end{split} $$ Substituting these expressions into the equation for $ |\nabla v|^2 $ above, we get $$ \begin{split} |\nabla u|^2 & = (\partial_r u_r)^2 + \frac{1}{r^2} \left( (r \partial_r u_\theta)^2 + (\partial_\theta u_r - u_\theta)^2 \right) + \frac{1}{r^4} (r(\partial_\theta u_\theta + u_r))^2 \\ & = (\partial_r u_r)^2 + (\partial_r u_\theta)^2 + \frac{1}{r^2} (\partial_\theta u_r - u_\theta)^2 + \frac{1}{r^2} (\partial_\theta u_\theta + u_r)^2. \end{split} $$

Edit 2: I found a divergence-free counterexample to the inequality using the stream function $ \psi(r,\theta) = r \cos{\theta} $ which results in the velocity function $$ u(r, \theta) = (u_r, u_\theta) = (-\sin{\theta}, -\cos{\theta}). $$ Substituting this into the original expression yields the inequality $ \cos^2(\theta) \leq 0 $ which does not hold for most points $ \theta \in [0,2\pi) $.

ben_dvs
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  • Did you use the formulae below to calculate $\nabla \mathbf{u}$? At the beginning of the proof of Theorem 4, it said that: We introduce polar coordinates $x_1 = r\cos \theta, x_2 = r\sin \theta$. The radial and angular components of $\mathbf{u}$ are related to the cartesian components by $$u_1 = u_r \cos \theta - u_{\theta}\sin \theta, \quad u_2 = u_r \sin\theta + u_\theta \cos\theta$$ (see next comment) – River Li Jun 05 '25 at 01:41
  • and the polar expressions for the derivatives of a function $p$ are $$\frac{\partial p}{\partial x_1} = \frac{\partial p}{\partial r} \cos\theta - \frac{\partial p}{\partial \theta}\frac{\sin\theta}{r}, \quad \frac{\partial p}{\partial x_2} = \frac{\partial p}{\partial r} \sin\theta + \frac{\partial p}{\partial \theta}\frac{\cos\theta}{r}.$$ – River Li Jun 05 '25 at 01:41
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    @RiverLi My derivation for $ |\nabla u|^2 $ was slightly different. I just added it to the original post for reference. – ben_dvs Jun 05 '25 at 04:16
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    I am not familiar with your added content. I used the formulae above to calculate $\nabla \cdot \mathbf{u}$ which gives the same result as the paper (also in your post, the divergence-free condition). I don't know if we can calculate $\nabla \mathbf{u}$ by the above formulae. – River Li Jun 05 '25 at 05:58
  • @RiverLi After some looking around, I found another derivation for $ \nabla \mathbf{u} $ in these lecture notes in section 2.8. If we ignore the $ z $ components and use the Frobenius inner product, a similar expression for $ |\nabla u|^2 $ can be found. I hope this helps. – ben_dvs Jun 05 '25 at 15:25
  • It gives the same result as yours. But if your calculation is right, then it seems we need some condition on the partial derivatives to reach $|\nabla \mathbf{u}|^2 \geq \left( \frac{1}{r} \partial_\theta u_r \right)^2$. – River Li Jun 06 '25 at 02:38
  • @RiverLi Maybe the divergence-free condition would be useful here? I can test some divergence-free functions to see if they hold or not. – ben_dvs Jun 07 '25 at 02:50
  • But you have checked the example $u(r, \theta) = (u_r, u_\theta) = (\theta, 1)$? – River Li Jun 07 '25 at 02:59
  • @RiverLi Yeah, for that example we have $ \partial_r u_r = \partial_r u_\theta = \partial_\theta u_\theta = 0 $ and $ \partial_\theta u_r = 1 $, so the inequality becomes $$ \frac{1}{r^2} \theta^2 + \frac{1}{r^2} - \frac{2}{r^2} \geq 0 $$ which doesn't hold for $ \theta \in (1,2\pi) $. But this example is not divergence-free, of course. – ben_dvs Jun 07 '25 at 14:37
  • So you obtained a counterexample? What conclusion can you draw? – River Li Jun 18 '25 at 12:45
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    @RiverLi Since the latest counterexample is divergence-free, the inequality must follow from some other condition on $ u $ if it is true. I'm going to try and find a different proof of this theorem to see if they use the same argument as Heywood. – ben_dvs Jun 19 '25 at 18:29

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